The equation of a circle $C$ is $x^2+y^2+14x+6y+42 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2+14x) + (y^2+6y) = -42$ $(x^2+14x+49) + (y^2+6y+9) = -42 + 49 + 9$ $(x+7)^{2} + (y+3)^{2} = 16 = 4^2$ Thus, $(h, k) = (-7, -3)$ and $r = 4$.